An Indian rubber cord L metre long and area o | vedclass

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Question

(B) Consider a small element of length $dx$ at a distance $x$ from the free end of the rubber cord.The weight of the portion below this element is $W

Vedclass · Accepted Answer

$L^2Dg/2E$

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(B) Consider a small element of length $dx$ at a distance $x$ from the free end of the rubber cord.The weight of the portion below this element is $W = (A \cdot x \cdot D) \cdot g$.The stress at this section is $\sigma = \frac{W}{A} = x D g$.By Hooke's Law,the strain is $\epsilon = \frac{\sigma}{E} = \frac{x D g}{E}$.The extension of the small element $dx$ is $dl = \epsilon dx = \frac{x D g}{E} dx$.To find the total extension $l$,we integrate from $x = 0$ to $x = L$:$l = \int_{0}^{L} \frac{D g}{E} x dx = \frac{D g}{E} \left[ \frac{x^2}{2} \right]_{0}^{L} = \frac{L^2 D g}{2 E}$.

An Indian rubber cord $L$ metre long and area of cross-section $A$ $m^2$ is suspended vertically. The density of rubber is $D$ $kg/m^3$ and Young's modulus of rubber is $E$ $N/m^2$. If the wire extends by $l$ metre under its own weight,then the extension $l$ is:

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